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The rms current in a \(47-\Omega\) resistor is 0.50 A. What is the peak value of the voltage across this resistor?

Short Answer

Expert verified
The peak voltage across the resistor is approximately 33.2 V.

Step by step solution

01

Understand the Relationship Between RMS and Peak Voltage

The root mean square (RMS) current value is related to the peak current value by the equation \( I_{peak} = I_{rms} \times \sqrt{2} \). For voltage, \( V_{rms} = \frac{V_{peak}}{\sqrt{2}} \), or equivalently, \( V_{peak} = V_{rms} \times \sqrt{2} \). This relationship arises from the fact that RMS values are used to represent equivalent DC values in AC circuits.
02

Calculate RMS Voltage

First, use Ohm's law to calculate the RMS voltage across the resistor. Ohm's law states \( V = I \times R \), where \( I \) is the current and \( R \) is the resistance. Here, \( I_{rms} = 0.50 \, A \) and \( R = 47 \, \Omega \). Therefore: \[ V_{rms} = 0.50 \, A \times 47 \, \Omega = 23.5 \, V \]
03

Calculate Peak Voltage

Now that we have the RMS voltage, we can find the peak voltage using the relationship \( V_{peak} = V_{rms} \times \sqrt{2} \). Substituting the RMS voltage we found: \[ V_{peak} = 23.5 \, V \times \sqrt{2} \approx 33.2 \, V \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
Ohm鈥檚 Law is a fundamental principle in electrical circuits, representing the relationship between voltage, current, and resistance. It is expressed in the simple equation: \[ V = I \times R \] where:
  • \( V \) is the voltage across the resistor,
  • \( I \) is the current flowing through the resistor, and
  • \( R \) is the resistance of the resistor.
Ohm's Law is particularly useful for calculating unknown values if two of the variables are known. In the context of AC circuits, this rule still applies for RMS values just like in DC scenarios. This law helps us justify and compute the RMS voltage when you know the RMS current and the resistance. Remembering this relationship simplifies finding voltage values in many electrical circuits. It鈥檚 key to solve problems related to both resistance and current in various formats.
AC Circuits
Alternating Current (AC) circuits are the backbone of modern electricity systems. Unlike Direct Current (DC) that flows steadily in one direction, AC alternates its direction periodically. Thus, various quantities in AC circuits like voltage and current periodically change their values.

RMS (Root Mean Square) values are crucial in AC circuits as they allow us to express AC quantities as equivalent DC values. This is essential because it provides a useful means of comparison between AC and DC. Calculating RMS values involves using formulas such as:
  • \( I_{rms} = \dfrac{I_{peak}}{\sqrt{2}} \)
  • \( V_{rms} = \dfrac{V_{peak}}{\sqrt{2}} \)
This adjustment ensures that equipment and components operate reliably under AC supply, even if the actual current and voltage are varying over time. In real-life AC applications, these calculations help avoid overloading circuits, ensuring consistent appliance performance.
Peak Voltage
The peak voltage in AC circuits represents the maximum value of voltage reached in one cycle. Understanding peak voltage is essential since many applications require knowledge of both RMS and peak values to ensure proper functionality and safety.

In our exercise scenario, we computed the peak voltage from the RMS voltage with the formula:\[ V_{peak} = V_{rms} \times \sqrt{2} \]This equation highlights an essential conversion factor, \( \sqrt{2} \), which bridges RMS values to peak values for sine waveforms common in AC circuits. The peak voltage indicates the highest voltage that can be experienced by electronic components, aiding in designing circuits against potential voltage spikes. Knowing this helps with the protection of equipment and maintaining the accuracy and precision of electrical readings and operations.

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Most popular questions from this chapter

Suppose that two capacitors \(\left(C_{1} \text { and } C_{2}\right)\) are connected in series. Show that the sum of the energies stored in these capacitors is equal to the energy stored in the equivalent capacitor. IHint: The energy stored in a capacitor can be expressed as \(q^{2} /(2 C) . J\)

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Multiple-Concept Example 9 reviews the concepts that are important to this problem. A light bulb is wired in series with a 144\(\Omega\) resistor, and they are connected across a 120.0 -V source. The power delivered to the light bulb is 23.4 \(\mathrm{W}\) . What is the resistance of the light bulb? Note that there are two possible answers.

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